From Biot-Savart Law to Divergence of B (1)

Transcription

1 From Bot-Savart Law to Dvergence of B (1) Let s prove that Bot-Savart gves us B (r ) = 0 for an arbtrary current densty. Frst take the dvergence of both sdes of Bot-Savart. The dervatve s wth respect to the measurement pont and the ntegral s wth respect to the source pont, so we we can take the dervatve nsde the ntegral: B (r ) = 10 7 λ dτ J ( r ) ˆ Next, apply the dentty from the cover of Grffths: ( a b ) = b ( a ) a ( b ) Usng the dentty, we get λ J ( r ) ˆ λ = ˆ J λ ( ) J ˆ The frst term s zero, because J r ( ) depends only on the source pont, and the dervatve s wth respect to the measurement pont.

2 From Bot-Savart Law to Dvergence of B (2) The curl n sphercal coordnates s v = 1 r snθ + 1 r + 1 r 1 snθ θ snθ v φ ( ) φ v θ φ v r r r v φ ( ) ( r r v θ ) θ v r φ ˆ θ ˆ We have v = 1 r 2 ˆ r wth only a radal component, so r ˆ v = 1 r 1 snθ φ 1 r 2 θ ˆ + 1 r θ 1 r 2 φ ˆ But 1 2 s ndependent of angle, so r φ 1 r 2 = θ Thus the second term s zero and we have B (r ) = 0 1 r 2 = 0

3 Component Notaton Vector notaton s a convenent shorthand, and s powerful enough for Newtonan mechancs and OK for electrostatcs But you probably notced last lecture when we demonstrated that the Bot-Savart law for magnetostatcs was equvalent to B = µ 0 J that we had to nvoke some denttes that are hard to remember and hard to nterpret. Component notaton s an alternatve to vector notaton that makes such problems more tractable, and also can be extended to more complcated problems. Bascally, we drop the arrow-notaton or bold-face-notaton, and ndcate vector components by subscrpts: a a Thus an equaton lke F = ma becomes F = ma (whch s really 3 equatons for the 3 components of the force and acceleraton n both cases). (Note that not every logcal groupng of 3 numbers s a vector. Temperature, pressure, and humdty are a convenent group, but aren t a vector s the physcs sense. Vector really means somethng that can be changes n a certan way when vewed from a rotated coordnate system).

4 Dot Products, Summaton Notaton A dot-product sn t a vector at all, but a scalar number wth no drecton. What t means s A B = A x B x + A y B y + A z B z ( ) or any other The subscrpt mght be ( x,y,z), or r,θ,φ three orthogonal drectons (but beware of rght vs left!) If we just use ( 1,2,3) for the ndces, we can make a more compact dot product: A B = 3 =1 A B Such sums come up so often that we ll assume summaton over any repeated ndex unless stated otherwse: A B = A B (The dea of assumed summaton over repeated ndces was nvented by Ensten, and he was as proud of that as he was of relatvty!)

5 Kronecker Delta Symbol It s useful to defne the Kronecker delta symbol δ j whch s unty f = j and zero otherwse: δ j = +1 f = j 0 f j The Kronecker delta symbol does for sumatons what the Drac delta functon does for ntegrals: A δ j = A δ j = A j We often run nto combnatons lke j A B j δ j = A B j δ j By totally conventonal algebra, we can swap the order of the summatons over and j, factor B j out of the sum over, then apply the prevous result: But ths s just A B! A B j δ j = B j A δ j = B j A j j j j

6 Dot Products and Matrx Notaton If we vew A and B as matrces wth 1 column and 3 rows, we could also wrte the dot-product as a matrx product: A B = A T B = B T A The superscrpt T means matrx transpose. If we wrte the components explctly, [ ] A B = A T B = A A 2 A 3 B 1 B 2 B 3 = A 1 B 1 + A 2 B 2 + A 3 B 3 Remember that wth matrces, the multplcaton-order matters: BA T = AB T s a 3 by 3 matrx, not a scalar! The components of δ j n matrx notaton are δ j = It s just the unt matrx!

7 Cross Products and Alternatng Tensor In Cartesan coordnates, the cross product s A B = x ˆ ( A y B z A y B z )+ y ˆ A z B x A x B z ( ) + ˆ ( ) z A x B y A y B x To wrte ths n component notaton compactly, we nvent a symbol called the alternatng tensor ε jk ε jk = 0 f = j, j = k, or k = +1 f jk =123, 231, f jk = 321, 213, 132 Thnk of the 3 ndces as wrappng around: j k j k... If the order s ncreasng, t s +1. If the order s decreasng, t s -1. If any two ndces are the same, t s zero. The k-th component of the cross product s: ( A B ) k = ε jk A B j j = ε jk A B j The frst ndex of ε jk goes wth A, the second wth B, the thrd s the ndex of the resultng cross-product vector. Let s check t. For the z-component of the cross-product, we want ( k = 3). Only the terms ( = 1, j = 2) and = 2, j =1 are non-zero, and these gve us ( +A 1 B 2 A 2 B 1 ) whch s just ( +A x B y A y B x ), whch s rght. ( )

8 Double-Cross-Products The double-cross-product dentty s A ( B C ) = B ( A C ) C ( A B ) ( ) mean? It says take the What does a term lke B A C vector B, and multply t by the scalar number ( A C ). Usng our new notaton we wrte [ A ( B C )] k = ε jk A ε lmj B l C m j The term n parentheses s the j component of B C, and we use l,m as the summaton ndces to make t clear they are dfferent from the,j ndces of the other summaton. Wth the sums explct lke ths, everythng s just a number and the order of operatons doesn t matter, so we wrte [ A ( B C )] k = ε jk ε lmj A B l C m jlm lm

9 Alternatng Tensor Identty If you can remember ths dentty, you can derve all the double-cross denttes on the fly: k ε jk ε lmk = δ l δ jm δ m δ jl For a gven k value, the two ε s wll have the same sgn f = l and j = m, and have the opposte sgn f = m and j = l, and any other combnaton wll gve zero. Note that the last ndex of the two ε s match. You can always get ths by usng the cyclc property of the ε s: ε jk = ε jk = ε kj The other 4 ndces could be anythng n any order; just transcrbe them to the δ s n the prescrbed pattern: a δ wth both frst ndces tmes a δ wth both second ndces, mnus the δ s wth the frst ndex from one and the second ndex from the other.

10 Double-Cross Revsted Apply ths to the double-cross product: [ A ( B C )] k = ε jk ε lmj A B l C m Frst, make the thrd ndces agree = ε kj ε lmj A B l C m jlm Next, use the sum over j to turn the product of ε s to a product of δ s: jlm = ( δ kl δ m δ km δ l )A B l C m lm Then, evaluate the δ s, rememberng that we expect to end up wth a k-ndex left over: ( ) = A B k C A B C k Then wrte the result n vector notaton agan: = B ( A C ) C ( A B )

11 Dervatves n Component Notaton The gradent of a scalar (whch s a functon of a vector) s f ( x ) = f x x ˆ + f y y ˆ + f z z ˆ In component notaton, the -th component s [ f ( x )] = f x Ths comes up so often that there s a shorthand verson: ( ) x Gradent: ( f ) f Dvergence: A A = A Curl: Note as well that ( A ) k ε jk A j x j = x x j = δ j j = ε jk A j

12 Provng Vector Dervatve Identtes Curl of cross-product, summatons assumed: [ ( A B )] k = ε jk ( ε lmj A l B m ) The ε s just a coeffcent, so t factors out of the dervatve. Then cycle the ndces on the frst ε: = ( ε kj ε lmj ) ( A l B m ) Now apply the alternatng-tensor dentty, and the product rule for the dervatve: Contract the δ s: = ( δ kl δ m δ km δ l )( A l B m + B m A l ) = ( A k B + B A k ) ( A B k + B k A ) We can now wrte t n vector notaton agan = A ( B ) + ( B )A ( A )B B ( A ) I fnd the meanng more clear n the component form!

13 Provng Vector Dervatve Identtes (2) Dvergence of a curl n component notaton: ( A ) = k ( ε jk A j ) Brng the dervatves together: = ε jk ( k A j ) But because k s symmetrc under k, but ε jk s antsymmetrc under k, ths has to be zero. The same argument works for the curl of a gradent: The curl of a curl: ( f ) = ε jk j f = 0 ( ) = ε jk ( ε lmj l A m ) = ε kj ε lmj l A m A ( ) = ( δ kl δ m δ km δ l )( l A m ) = ( k A ) ( A k ) = k ( A ) ( )A k Ths form s clearer than the equvalent vector form: = ( A ) 2 A

14 Vector Dervatves of Radal Vector Functons A radal vector functon ponts n the radal drecton, and s a functon only of radus. Whle we mght wrte t wth a radal unt vector and a functon of the magntude of the radus, we could equally well wrte t as x f ( x x ) In component form, usng two dfferent subscrpts x f j x j x j The vector dervatve operator gves x x f x j x j k j = f x j x j x x + x k j = f ( x 2 )δ k + x x k ( ) x 2 = f ( x 2 f x2 )δ k + 2 j ( ) x x k x k f x j x j Ths s symmetrc wth respect to k j f ( x2 ) ( x 2 ) x j x j

15 Bot-Savart to Dvergence of B, Revsted Start wth the Bot-Savart ntegral, usng r for the source coordnate and r for the measurement pont B (r ) = 10 7 J r λ λ d τ λ = r r 2 ( ) ˆ Let s take the dvergence of B n component notaton k B k = 10 7 k ε jk J r λ j ( ) ˆ d τ Move the through the ε, and through the J ( r ), snce the dervatve s wth respect to the measurement pont rather than the source pont: = 10 7 J r λ j ( ) ε jk k ˆ λ d τ 2 Ths dervatve wll be symmetrc under exchange of j and k, but the ε s antsymmetrc, so the result s zero. Wasn t that smpler than the vector-notaton verson?

16 Now the curl of B: Curl of B, Revsted ε jk B j =10 7 ε jk ε lmj J l r ( ) ˆ λ m d τ Move the ε s together, move the through the J ( r ): = 10 7 ε jk ε lmj J l λ ˆ m λ d τ 2 Cycle the ndces and apply the ε δ dentty: Contract the δ s = 10 7 ε kj ε lmj J l λ ˆ m λ d τ 2 = 10 7 ( δ kl δ m δ km δ l )J l λ ˆ m λ d τ 2 = 10 7 J k λ ˆ λ J 2 ˆ λ k d τ

17 Curl of B, Revsted (2) Do the second term frst, usng some trcks. Frst, by the defnton λ = r r, we can change dervatve varables by changng the sgn: J λ ˆ k λ = J λ ˆ 2 k Next, from the dervatve of products rule we can say J J ˆ λ k ˆ λ k = λ ˆ k J + J = λ ˆ k J J ˆ λ k ˆ λ k The left term s zero for magnetostatcs wth steady currents because J = J = 0. The rght term s a pure dvergence, and we are ntegratng over t. We expand the ntegraton volume untl there are no currents at the surface, and apply the dvergence theorem: J λ ˆ k d τ = J λ ˆ k d a = 0 surface volume

18 Curl of B, Revsted (3) Now the remanng term that we skpped over before ε jk B j =10 7 J k λ ˆ ( r ) λ d τ 2 Translatng back nto vector form, we recognze as our frend the 3-d Drac delta functon ˆ λ = ˆ λ = 4πδ3 λ ( ) Thus we get ε jk B j =10 7 J k 4πδ ( 3 λ )d τ = 4π 10 7 J k ( r ) Back n vector notaton, ths s B = µ 0 J